c++ - 当没有调用“this”时如何修复“this was nullptr”
问题描述
我用一个简单的ai写了一个简单的tictactoe游戏。初始化我的游戏对象时出现错误。我的游戏对象使用名为 setPlayerNum 的方法初始化两个玩家对象,每个对象的玩家编号属性设置为“一”或“二”。尽管我从不使用“this”关键字,但我收到一条错误消息,指出“这是一个 nullptr”以及写访问冲突。
我已经尝试取消引用播放器对象以及将播放器编号属性设为公共变量(以防它是范围错误)。都不适合我。
int main() {
game g;
g.printTitleCard();
std::system("pause");
return 0;
}
#include "board.h"
#include "player.h"
class game {
public:
game();
~game();
void printTitleCard();
void play();
int victory();
void victoryScreen(int winner);
private:
board * gameBoard;
player * player_one;
player * player_two;
int turns;
int maxTurns = 9;
};
game::game() {
gameBoard = new board();
player_one->setPlayerNum(1);
player_two->setPlayerNum(2);
}
class player {
public:
player();
~player();
void setPlayerNum(int);
int getPlayerNum();
void updateScore(int);
int retrieveWins();
int retrieveLoses();
int retrieveTies();
private:
int playerNum = 0;
int wins = 0;
int loses = 0;
int ties = 0;
};
void player::setPlayerNum(int num) {
playerNum = num;
}
解决方案
game::game() {
gameBoard = new board();
player_one->setPlayerNum(1);
player_two->setPlayerNum(2);
}
在调用之前不实例化任何player
s setPlayerNum
,这意味着没有有效的实例player
可以调用setPlayerNum
。在此之后,程序陷入了未定义的行为,所有的赌注都被取消了。
解决方案:
解决此问题的最佳方法是返回几个步骤
class game {
public:
game();
~game();
void printTitleCard();
void play();
int victory();
void victoryScreen(int winner);
private:
board * gameBoard;
player * player_one;
player * player_two;
int turns;
int maxTurns = 9;
};
gameBoard
并且不要对,player_one
和使用指针player_two
。
board * gameBoard;
player * player_one;
player * player_two;
变成
board gameBoard;
player player_one;
player player_two;
和
game::game() {
gameBoard = new board();
player_one->setPlayerNum(1);
player_two->setPlayerNum(2);
}
变成
game::game() {
player_one.setPlayerNum(1);
player_two.setPlayerNum(2);
}
仅new
在绝对必要时使用,并且在几乎从未有过的库容器和智能指针的世界中使用。更多信息:为什么 C++ 程序员应该尽量减少“新”的使用?
无关:
如果您更改播放器构造函数以将播放器编号作为参数,setPlayerNum
则可能变得不必要并且game
' 构造函数可以变为
game::game(): player_one(1), player_two(2){
}
附录
有什么办法可以将它们保留为指针?
是的,但这不是一个好主意,因为你要承担大量额外的责任。new
在使用它们之前,您在构造函数中对播放器进行导航。您必须delete
在播放器和game
析构函数中,并且必须编写自己的副本(如果需要,可以移动)构造函数和赋值(以及移动赋值)运算符,以确保正确复制和分配对象(请参阅什么是三法则?更多关于你为什么需要这些东西的信息)。
game::game() {
gameBoard = new board();
player_one = new player; //added to get player instance
player_two= new player; // added to get player instance
player_one->setPlayerNum(1);
player_two->setPlayerNum(2);
}
// new function: copy constructor
game::game(const game & source) {
gameBoard = new board(*source.gameBoard);
player_one = new player(*source.player_one);
player_two= new player(*source.player_two);
turns = source.turns;
maxTurns = source.maxTurns;
}
game::~game() {
delete gameBoard; // you needed to have this if you didn't already
delete player_one; // added to release player instance
delete player_two; //added to release player instance
}
// new function swap function (used by assignment operator)
game::swap(game & source) {
std::swap(gameBoard, source.gameBoard);
std::swap(player_one , source.player_one);
std::swap(player_two, source.player_two);
std::swap(turns, source.turns);
std::swap(maxTurns, source.maxTurns);
}
// new function assignment operator
game & operator=(game source)
{
swap(source);
return * this;
}
看看你必须写的所有额外的东西。很可能那里有一两个错误。
赋值运算符正在利用Copy and Swap Idiom。这通常不是最快的分配方式,但它非常容易编写并且很难出错。我从复制和交换开始并使用它,直到生活证明它是性能瓶颈。
如果 gameBoard
和是自动变量player_one
,player_two
编译器将为您生成析构函数和所有复制代码,除非board
写得不好。您的目标之一应该是让拥有资源的所有类都遵守三或五规则,以便包含它们的类可以利用零规则。